Question

If $x-\frac{1}{x}=-\sqrt{3}$, then find $x^3-\frac{1}{x^3}$.

• A. $6\sqrt{3}$
• B. $2\sqrt{3}$
• C. $3\sqrt{3}$
• D. $-6\sqrt{3}$

Answer 1

The correct option is D $-6\sqrt{3}$

$x-\frac{1}{x}=-\sqrt{3}$

Take cube of both sides, we get

${(x-\frac{1}{x})}^3={(-\sqrt{3})}^3$

$\Rightarrow x^3-\frac{1}{x^3}-3\cdot x\cdot \frac{1}{x}(x-\frac{1}{x})=-3\sqrt{3}\text{by using}(a-b{)}^3=a^3-b^3-3ab(a-b)$

$\Rightarrow x^3-\frac{1}{x^3}-3(-\sqrt{3})=-3\sqrt{3}$

$\Rightarrow x^3-\frac{1}{x^3}+3\sqrt{3}=-3\sqrt{3}$

$\Rightarrow x^3-\frac{1}{x^3}=-6\sqrt{3}$

So, option $D$ is correct.